Question

32^[(x+5)/(x-7)] = (.25)(128^[(x+17)/(x+3)]

Answer 1

32^[(x+5)/(x-7)] = (.25)(128^[(x+17)/(x+3)]


32^[(x+5)/(x-7)] = (1/4)(128^[(x+17)/(x+3)]


(2^5)^[(x+5)/(x-7)] = 2^(-2)((2^7)^[(x+17)/(x+3)]


2^[5(x+5)/(x-7)] = 2^(-2)(2^[7(x+17)/(x+3)]


2^[5(x+5)/(x-7)] = 2^[-2+(7(x+17))/(x+3)]


Since the bases are equal (to 2), the exponents must be equal, so

5(x+5)/(x-7) = -2+(7(x+17))/(x+3)

Solve that for x to find your answer

Answer 2

I got that far. The answer is supposed to be 10, but I didn't get 10. I got some weird fraction. Is it right to first combine the terms on the right by finding the common denominator of x + 3?

Answer 3

hmm I'm getting a fraction too, the original equation is this right


\[\Large 32^{\frac{x+5}{x-7}} = (0.25)\left( 128^{\frac{x+17}{x+3}}\right ) \]


???

Answer 4

yes

Answer 5

what fraction did you get for the answer?

Answer 6

I don't remember but it was something about dividing 68 into something or something like that. I erased what I wrote

Answer 7

I guess it doesn't matter, but I'm getting 433/19 for the answer. Not sure how they got 10.

Answer 8

let me check 10

Answer 9

ok

Answer 10

it doesn't check i don't think...

Answer 11

you're right, it is 433 / 19.
http://www.wolframalpha.com/input/?i=32%5E%5B%28x%2B5%29%2F%28x-7%29%5D+%3D+%28.25%29%28128%5E%5B%28x%2B17%29%2F%28x%2B3%29%5D+

Answer 12

hmm must be a typo somewhere, who knows

Answer 13

probably, thanks anyway

Answer 14

np

Answer 15

can i message you later if I need help with another problem?

Answer 16

sure

Answer 17

[(x+17)/(x+3)]

(x+17)/(x-3)

Answer 18

hmm seems too easy, thanks Algebraic!

Answer 19

32^[(x+5)/(x-7)] = (.25)(128^[(x+17)/(x-3)]

32^[(x+5)/(x-7)] = (1/4)(128^[(x+17)/(x-3)]

(2^5)^[(x+5)/(x-7)] = 2^(-2)((2^7)^[(x+17)/(x-3)]

2^[5(x+5)/(x-7)] = 2^(-2)(2^[7(x+17)/(x-3)]

2^[5(x+5)/(x-7)] = 2^[-2+(7(x+17))/(x-3)]

Since the bases are equal (to 2), the exponents must be equal, so

5(x+5)/(x-7) = -2+(7(x+17))/(x-3)

Solve that for x to find your answer